&= for bools?

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proxyuser

I see the equivalent of this in some code.

bool a, b, c;
a &= b;

Is that correct? I would have thought it would be
a &&= b;

Of course I'm not saying &&= is an operator, but

a = a && b;
 
I see the equivalent of this in some code.

bool a, b, c;
a &= b;

Is that correct?  I would have thought it would be
a &&= b;

Of course I'm not saying &&= is an operator, but

a = a && b;
Click to expand...

According to the help library, &= is a valid assignment operator, but
&&= is not listed.
 
I see the equivalent of this in some code.

bool a, b, c;
a &= b;

Is that correct? I would have thought it would be
a &&= b;

Of course I'm not saying &&= is an operator, but

a = a && b;
Click to expand...

I guess what's happening behind the scenes is that the bools are being
treated as the numbers they actually are and their bits are being AND'ed. Of
course, since there are only two possible underlying values for a bool,
AND'ing them will either result in the True value if they're both True or
the False value otherwise, so the ultimate result will be the same as
applying the && operator.
 
According to the help library, &= is a valid assignment operator, but
&&= is not listed.
Click to expand...

More to the point (from the entry):

"The & operator performs a bitwise logical AND operation on integral
operands and logical AND on bool operands."
 
Geoffrey Summerhayes said:
More to the point (from the entry):

"The & operator performs a bitwise logical AND operation on integral
operands and logical AND on bool operands."
Click to expand...

Ah, I missed that, thanks. Does this beg the question "Why is the &&
operator needed then?" ?
 
Ah, I missed that, thanks.  Does this beg the question "Why is the &&
operator needed then?" ?
Click to expand...

It's not. If a is true, b does not get evaluated with &&, so it can
save time.

&& is mainly used in cases when you need to check an object's property
and you are not sure the object has been instantiated yet, as in:

if ((obj != null) && (obj.foo == "ABC"))
 
Ah, I missed that, thanks.  Does this beg the question "Why is the &&
operator needed then?" ?
Click to expand...

&& is a logical operator, and & is the bitwise operator. The
expression on each side of && MUST evaluate to a boolean. & can be
used with other number types.
 
&& is a logical operator, and & is the bitwise operator.  The
expression on each side of && MUST evaluate to a boolean.  & can be
used with other number types.
Click to expand...

Please read the thread to which you're responding before doing that.
As noted by Geoffrey, in C# what you say is wrong - operators & and |
are Boolean and not bitwise when applied to bool operands (and yes,
they can be used with bool).
 
Ah, I missed that, thanks.  Does this beg the question "Why is the &&
operator needed then?" ?
Click to expand...

&& and || are short-circuiting (do not evaluate 2nd operator if the
result of expression is evident from the first one). & and | aren't.
Sometimes, when you have side-effects in a second operand, you care
about the difference, though in practice it is so rare that doing so
is potentially very confusing. VB has a similar distinction between
And/Or and AndAlso/OrElse.
 
It's not. If a is true, b does not get evaluated with &&, so it can
save time.

&& is mainly used in cases when you need to check an object's property
and you are not sure the object has been instantiated yet, as in:

if ((obj != null) && (obj.foo == "ABC"))
Click to expand...

I assume you meant to write "If a is false...", right? I wonder why this
same optimization couldn't be implemented for &.
 
Pavel Minaev said:
&& and || are short-circuiting (do not evaluate 2nd operator if the
result of expression is evident from the first one). & and | aren't.
Sometimes, when you have side-effects in a second operand, you care
about the difference, though in practice it is so rare that doing so
is potentially very confusing.
Click to expand...

I see, thanks.
 
proxyuser said:
I see the equivalent of this in some code.

bool a, b, c;
a &= b;

Is that correct? I would have thought it would be
a &&= b;

Of course I'm not saying &&= is an operator, but

a = a && b;
Click to expand...

The && operator does a short circuit evaluation, which means that if the
first operand is false the second operand doesn't need to be evaluated.

For an assigment operator short circuiting is not really useful. The &&=
operator would be the equivalent of:

if (a) a &= b;

That means that a short circuit assignment operator would actually do
more work than the regular assignment operator in some cases.

If you know that there is a lot of work to evalue the operand, you can
just slap an if in front of the assignment operator.
 
&& and || are short-circuiting (do not evaluate 2nd operator if the
result of expression is evident from the first one). & and | aren't.
Sometimes, when you have side-effects in a second operand, you care
about the difference, though in practice it is so rare that doing so
is potentially very confusing. VB has a similar distinction between
And/Or and AndAlso/OrElse.
Click to expand...

I don't find it rare or confusing. I use lazy evaluation quite often. Some
examples:

if (!String.IsNullOrEmpty(myString) && myString.Length == 5) //if myString
were null the length check would throw an exception.

if (File.Exists(...) || DownloadFile(...)) //Useful for caching. If file is
available or it's sucessfully downloaded proceed.

I admit that if you aren't careful things can get confusing, however, that
statement applies to virtually any language feature.

Andrew Faust
 
I don't find it rare or confusing. I use lazy evaluation quite often. Some
examples:

if (!String.IsNullOrEmpty(myString) && myString.Length == 5) //if myString
were null the length check would throw an exception.

if (File.Exists(...) || DownloadFile(...)) //Useful for caching. If file is
available or it's sucessfully downloaded proceed.
Click to expand...

I meant it the other way around. Virtually always, you _want_ short
circuiting, which is why many languages out there don't even have non-
short-circuiting boolean operators. To that end, I'm saying that using
a non-short-circuiting operator because you need to rely on a side
effect is a second expression is something that occurs so rarely that
it may confuse a typical reader of the code.
 
For an assigment operator short circuiting is not really useful. The &&=
operator would be the equivalent of:

if (a) a &= b;

That means that a short circuit assignment operator would actually do
more work than the regular assignment operator in some cases.
Click to expand...

How so? An expansion of the non-short-circuiting operator also have to
do the same thing (i.e. check the value of "a" before assigning a new
one), because that's just how logical AND and OR work. The only
difference would be that a short-circuiting form would not evaluate b
if it's not needed.
 
Pavel said:
How so? An expansion of the non-short-circuiting operator also have to
do the same thing (i.e. check the value of "a" before assigning a new
one), because that's just how logical AND and OR work. The only
difference would be that a short-circuiting form would not evaluate b
if it's not needed.
Click to expand...

No, the non-short-circuiting operator doesn't check the vale of a before
doing the operattion.

This code:

a &= b;

compiles into:

movzx eax, byte ptr [rsp+20h]
movzx ecx, bytr ptr [rps+21h]
and eax, ecx
mov byte ptr [rsp+20h], al

A short-circuiting operator would have to make a test on the value from
a, then make a conditional jump depending on the result:

movzx eax, byte ptr [rsp+20h]
test eax, eax
je .skip
movzx ecx, bytr ptr [rps+21h]
and eax, ecx
mov byte ptr [rsp+20h], al
..skip
 
No, the non-short-circuiting operator doesn't check the vale of a before
doing the operattion.

This code:

a &= b;

compiles into:

movzx eax, byte ptr [rsp+20h]
movzx ecx, bytr ptr [rps+21h]
and eax, ecx
mov byte ptr [rsp+20h], al
Click to expand...

I was assuming that "a" and "b" are placeholders for arbitrary
expressions, not just variables. I see that it used the AND operator
here - but I doubt whether it will do the same in e.g. a statement
such as:

foo.Bar.Baz &= (x > y);

I'd rather expect it to stick to conditional jumps here (since it'll
have to do CMP either way, might as well jump right then, instead of
doing SET/AND).

By the way, the above disassembly is from the x64 JIT, right? Have you
tried it on x86? the difference between implementations can be very
drastic at times.
 
Pavel said:
No, the non-short-circuiting operator doesn't check the vale of a before
doing the operattion.

This code:

a &= b;

compiles into:

movzx eax, byte ptr [rsp+20h]
movzx ecx, bytr ptr [rps+21h]
and eax, ecx
mov byte ptr [rsp+20h], al
Click to expand...

I was assuming that "a" and "b" are placeholders for arbitrary
expressions, not just variables. I see that it used the AND operator
here - but I doubt whether it will do the same in e.g. a statement
such as:

foo.Bar.Baz &= (x > y);
Click to expand...

Yes, it does it the same way. There is more code to get the location of
foo.Bar.Baz and to evaluate the expression x > y, but the code for the
&= operation is still an and operation:

; get foo.Bar
mov rax,qword ptr [rsp+30h]
cmp byte ptr [rax],0
mov rcx,qword ptr [rsp+30h]
call FFFFFFFFFFEC96D0
mov qword ptr [rsp+40h],rax

; get Baz
mov rax,qword ptr [rsp+40h]
movzx edx,byte ptr [rax+8]

; evaluate x > y
xor ecx,ecx
mov eax,dword ptr [rsp+28h]
cmp dword ptr [rsp+24h],eax
setg cl
mov dword ptr [rsp+48h],ecx

; &=
mov ecx,edx
and ecx,dword ptr [rsp+48h]

; set Baz
mov rax,qword ptr [rsp+40h]
mov byte ptr [rax+8],cl
I'd rather expect it to stick to conditional jumps here (since it'll
have to do CMP either way, might as well jump right then, instead of
doing SET/AND).
Click to expand...

Nope, no jumps. :)
By the way, the above disassembly is from the x64 JIT, right? Have you
tried it on x86? the difference between implementations can be very
drastic at times.
Click to expand...

Good point. Yes, it was x64 code. The x86 code looks like this:

mov eax, dword ptr [ebp-28h]
and eax, dword ptr [ebp-2Ch]
and eax, 0ffh
mov dword ptr [ebp-28h]

It reads and writes the bool values as dwords instead of bytes, but
other than that it's pretty much the same.
 
Pavel said:
No, the non-short-circuiting operator doesn't check the vale of a before
doing the operattion.
This code:
a &= b;
compiles into:
movzx eax, byte ptr [rsp+20h]
movzx ecx, bytr ptr [rps+21h]
and eax, ecx
mov byte ptr [rsp+20h], al
Click to expand...
Click to expand...
I was assuming that "a" and "b" are placeholders for arbitrary
expressions, not just variables. I see that it used the AND operator
here - but I doubt whether it will do the same in e.g. a statement
such as:
Click to expand...
   foo.Bar.Baz &= (x > y);
Click to expand...

Yes, it does it the same way. There is more code to get the location of
foo.Bar.Baz and to evaluate the expression x > y, but the code for the
&= operation is still an and operation:

; get foo.Bar
mov rax,qword ptr [rsp+30h]
cmp byte ptr [rax],0
mov rcx,qword ptr [rsp+30h]
call FFFFFFFFFFEC96D0
mov qword ptr [rsp+40h],rax

; get Baz
mov rax,qword ptr [rsp+40h]
movzx edx,byte ptr [rax+8]

; evaluate x > y
xor ecx,ecx
mov eax,dword ptr [rsp+28h]
cmp dword ptr [rsp+24h],eax
setg cl
mov dword ptr [rsp+48h],ecx

; &=
mov ecx,edx
and ecx,dword ptr [rsp+48h]

; set Baz
mov rax,qword ptr [rsp+40h]
mov byte ptr [rax+8],cl
I'd rather expect it to stick to conditional jumps here (since it'll
have to do CMP either way, might as well jump right then, instead of
doing SET/AND).
Click to expand...

Nope, no jumps. :)
By the way, the above disassembly is from the x64 JIT, right? Have you
tried it on x86? the difference between implementations can be very
drastic at times.
Click to expand...

Good point. Yes, it was x64 code. The x86 code looks like this:

mov eax, dword ptr [ebp-28h]
and eax, dword ptr [ebp-2Ch]
and eax, 0ffh
mov dword ptr [ebp-28h]

It reads and writes the bool values as dwords instead of bytes, but
other than that it's pretty much the same.
Click to expand...

Certainly very interesting. Another question, and I realise it may be
a dumb one, but did you make sure that JIT optimization wasn't
disabled because the debugger is attached? (as you surely know, the
default JIT behavior is to disable optimizations when running under
debugger).

In any case, getting back to why we started talking about it - a short-
circuiting &&= would be useful in precisely the same situation where a
short-circuiting && is useful today - when the second argument is
potentially too expensive to be evaluated. In practice, more often
than not it's a reasonable default - if the right side of && is a
local variable, then the compiler can do whatever it wants (e.g. AND),
since "evaluating" the variable doesn't have any visible side effects
as far as programmer is concerned; and if there are any potential side
effects (as is the case for field access and method calls), then it is
most likely that evaluating the right side will take more time then
doing a TEST/JE on the left side to short-circuit. This all equally
applies to &&=, naturally, so it would be reasonable to have it in the
language, and reasonable to use it by default for most cases.
 
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